The Physics Philes, lesson 30: Return of Physics!

In which tennis balls are hit, momentum is gained, and physics is resurrected.

Lately I’ve been doing a lot of math in this small corner of the Internet. Well, trying to, anyway. But then I realized, hey! This is The Physics Philes, not the Math…Thingy. So this week we see the return of physics! Hopefully I haven’t forgotten too much. One thing I noticed after so many weeks of working on strictly math is that the physics book I’m using is much, much more stringent than the book I’m using for calculus. It’s taking some time to adjust. But I’ll never get used to it if I don’t jump in head first, right? So let’s go!

This week we’ll be looking at the related topics of momentum and impulse.

First, as with any new topic, we need to define what we’re talking about. Momentum is a vector quantity. It’s magnitude is the product of an object’s mass and velocity. Impulse is basically a force that’s applied to a body for a certain amount of time. As equations, they look like this:

Momentum:

Impulse:

As an aside, I’d love to explain for you how the momentum equation is derived from Newton’s second law of motion, but it seems to use derivatives and I don’t know how those work yet. So I don’t think I have a lot of room to speak on the issue. But someday! Yes…someday…

*ahem* Anyhoo. Let’s step back a moment and just focus on momentum. A rapid change in momentum needs a large net force. A gradual change in momentum needs a smaller net force. The equation for momentum may look a little familiar. It kind of looks like the equation for kinetic energy (K = 1/2mv^2). The fundamental difference between these two are that momentum is a vector whose magnitude is proportional to an object’s speed. Kinetic energy, on the other hand, is a scalar quantity that is proportional to speed squared. But what does that look like in the real world? That is where impulse comes in.

As we mentioned before, impulse is the product of the net force and the time interval involved. (Assuming, of course, that we’re talking about a constant net force.) When an object receives an impulse it undergoes a change in velocity, and the change in velocity depends on the mass of the object. As you can tell by the equation above, impulse is a vector quantity. It’s direction is the same direction as the net force.

If you’re like me, you’re finding this hard to visualize. Just think of hitting a baseball with a bat, or a tennis ball with a tennis racket. The force you apply to the ball via the bat or racket during the time the bat or racket is in contact with the ball is the impulse.

As you can probably tell at this point, momentum and impulse can be intimately connected. In fact, there is a theorem where the two words are separated by a mere hyphen: the impulse-momentum theorem.

To explain the impulse-momentum theorem, one needs to know what Newton’s second law of motion looks like in terms of momentum. I can tell you that, but I can’t explain why. This is the best I can do right now.

Newton’s second law of motion in terms of momentum looks like this:

Again, please don’t ask me why. If the net force is consistent, then dp/dt is also consistent and equal to the total change in momentum Δp during Δt, divided by Δt. Which looks like this in math:

If we multiply each side by the denominator we get:

Oh lookie. If you’ll remember from waaaay at the top of this post, impulse equals the net force times the time interval. Which means that impulse is also equal to change in momentum!

Ta-da! The impulse-momentum theorem! The change in the momentum of a particle during a time interval equals the impulse of the net force that acts on the particle during that interval. Now that we know the equation, next week we’ll work some problems so we can see what momentum and impulse might look like in meatspace.

In the meantime, please leave any corrections or clarifications in the comments. I feel like my grasp of this subject is a little tenuous so far, so anything you can do to solidity it would be greatly appreciated.

1 Comment

You haven’t got to derivatives yet in the calc book, but the concept isn’t that hard if you’re not going for full rigor. A derivative of something is that thing’s rate of change. So the derivative of ‘p’ with respect to ‘t’ (i e dp/dt) is how fast the quantity p is changing with respect to changes in t.

(You need to specify the with-respect-to part, because your speed will be different numbers if you’re giving it in miles-per-hour or in miles-per-second. Also it will be zero in miles-per-apple, since changing the number of apples you have doesn’t change your position.)

The standard first physical example of derivatives in a calc class is: velocity is the derivative of position with resp. to time. The second is: acceleration is the derivative of velocity.

(So: velocity is how fast position is changing, hence measured in m/s. Acceleration is how fast velocity is changing, hence (m/s)/s)

Now, derivatives don’t mess with constant multiples in much the same way limits don’t. Example: As soon as you know that there are 1.6 km in a mile (OK, approximately) you also know that your speed in km/hour is 1.6 times your speed in mph. To write the math notation: K = 1.6*M implies that (dK/dt)=1.6*(dM/dt).

(K and M are your position in km and miles respectively.)

So: since mass isn’t changing, knowing that P=mV (I don’t have arrows so I’m just capitalizing the vectors) you also know that (dP/dt) = m*(dV/dt), and (dV/dt) is the acceleration. Therefore F=ma comes by way of: F = (dP/dt) = m(dV/dt) = ma.

(The sigma is a summation sign, right? So that’s just the total force on the object, i e the sum of all the forces?)

Now the theorem: when you say consistent you mean that the force and dP/dt are constant, right? If a rate of change is constant then its the same as the average rate of change, which is much easier to compute. An average rate of change (w.r.t time) is just the quotient: (Total change)/(Change in time).

(If I go 40 miles in 2 hours, what was the velocity? Note that, if I stopped off for a hot dog for the middle hour, my average velocity will still be 20mph, so I’ll have to drive 40mph in the first and last half-hours to make up the difference. Probably more than that due to stop lights. If I was on the highway, and have cruise control, I can have constant velocity, which then must be 20mph the whole time, or I’d go further. Assume I somehow don’t get pulled over or cause an accident.)